Hi there folks, I just wanted to see if I could get a quick clarification on the way damage is dealt in the final battle.

The rule in the rulebook is as follows:

Unlike a normal battle, the Ancient One cannot be defeated in a single attack. Instead, keep track of every success an investigator scores against the Ancient One. These successes are cumulative, and each successive player adds to them with his own attack. When the players accumulate a total number of successes equal to the number of players (including any players that were eliminated from the game), remove one doom token from the Ancient One’s doom track and reset their cumulative successes to zero.

The way I've interpreted this is that each player rolls all their attack dice, but can do a maximum of 1 damage per round, because the of the cumulative success wipe.

So, for example, a game with 4 players:

• player one rolls 8 dice, three successes. No damage is dealt.
• player two rolls 6 dice, five successes. The first success combines with the three existing, and one damage is dealt to the Ancient One.
• the other four successes are ignored because of the reset, and play moves on.

This seems to me to be what the rules intend, but this is a ruling by the game designer:

Defeating the Ancient One (07/09/05)
Basically, to defeat the Ancient One, the players must do a total # of successes equal to (# of players) x (# of doom tokens on Ancient One). So, if 4 players are facing Yig (doom track of 10), they need 40 successes to win. For every 4 successes they do, they remove 1 doom token to track their progress.

This seems to ignore the reset mentioned in the rulebook. Does anyone have a firm example of how this is meant to work?

Also, just to be certain, cumulative successes carry over between player rounds, correct?

Thanks all!

This is a very common question because what the rulebook says is just wrong!

Successes are not "reset to zero" when you remove a doom token, so any excess successes can help to remove the next doom token.

Thanks for the quick reply. I wonder what the intent of that reset in the rules is then…

I think it's just a clumsily phrased explanation.

Rulebook says:

"Unlike a normal battle, the Ancient One cannot be defeated in a single attack. Instead, keep track of every success an investigator scores against the Ancient One (see “Skill Checks,” page 13). These successes are cumulative, and each successive player adds to them with his own attack. When the players accumulate a total number of successes equal to the number of players (including any players that were eliminated from the game), remove one doom token from the Ancient One’s doom track and reset their cumulative successes to zero"

Depending on how you interpret that, it is SORT OF correct. It's meant to indicate that you're keeping a running total solely for comparison with the number needed to remove the next doom token . The thing you're meant to reset to zero is your count of "how many of the X successes required for this Doom token have we got so far". However, it's just the 'count' that you're resetting; you're not meant to ignore an actual dice. But yeah, it's such a subtle distinction that it's not a surprise that every single new player misunderstands it.

I really should just start copying and pasting this answer, instead of writing it out again every time someone asks…

So… I might be interpreting this entirely wrong, but is possible then for more than one doom token to be removed from an AO doom track per turn?

So, for example, a game with 4 players:

• player one rolls 8 dice, five successes. Remove a doom token?
• player two rolls 6 dice, five successes. Remove another doom token?
• player three rolls 5 dice, two successes. The two left over success from the previous rollers add to his, so a third doom token is removed?

Or in this case, the reset clause comes into play, and each time a doom token is removed form the track, the count of successes begins again, and player three's two successes are added to player four's rolls, and a third doom token can be removed if player four rolls an additional two successes?

Regardless, I've been doing this entirely wrong. I've only played two solo games thus far, and for whatever reason (me being a dummy probably ranks among the top) I've been counting each attack that has a 5 or 6 among the dice (regardless of multiple fives or sixes) as a single success, but that success adding a die to the next investigator's combat check pool. Then, if all investigators had a success in their attack, then a doom token was removed from the track.

I honestly couldn't tell you why I interpreted the rules that way, now.

Urgh. Obviously I need to go back to my organic chemistry, because I get less confused by that.

And "successes" is a funny word.

Nun of your business said:

What you wrote is totally correct.

I've never understood what the confusion over the wording in the rulebook is all about. It is plainly NOT wrong.

Example: 4-player game

1. First player rolls the dice, gets 2 successes.

2. Second player gets 2 more, now you have 4 total successes.

3. Remove a doom token.

4. Reset the count!

5. Third player gets 1 success.

6. Fourth player, no successes.

7. AO attacks.

8. Next turn!

9. First player gets 3 successes, now you have 4 total successes!

10. Repeat from point 3.

Tox said:

I think it's possibly even more sad that I never really disputed the wording, but just misinterpreted the entire thing. Cheers to a second grade reading comprehension on my part.

Thanks so much for the confirmation! I'm having some friends over tonight for our first group session, and thanks to the forums and all of your help, I think any rules akwardness should be all taken care of by the time our first gate opens.

Now if I could only stop my dice from always rolling "twos".